Beyond the Long Wavelength Limit

Beyond the Long Wavelength Limit Thus far, we have studied EM radiation by oscillating charges and current confined to a volume of linear size much smaller than the wavelength λ = πc/ω. In these notes, we shall work out EM radiation by an antenna of length comparable to λ or even longer than λ. But let s start with the general theory. For any harmonically oscillating current J(x, t) = J(x) exp( iωt), the vector potential of the radiation emitted by this current is(in the Landau gauge) A(x,t) = µ 0 e iωt d 3 y exp(ik x y ) x y This formula is valid for all x, but let s focus on the far zone where J(y). (1) r = x λ and r antenna size. () Note: for a small antenna it s the first condition here which defines the far zone, but for a large antenna the second condition takes over. And thanks to this second condition, we have x y for all y contributing to the integral (1), hence and therefore exp(ik x y ) x y x y r n y (3) exp(ikr) r Consequently, in the far zone we have a spherically divergent EM wave where A(x,t) µ 0 exp(ikr iωt) r f(n) = 1 exp( ikn y). (4) f(n), (5) d 4 yj(y)exp( ikn y). (6) The electric and the magentic fields in the far zone and hence the EM power radiated in the direction n follow from eq. (5) in the usual way, just as we learned last week: H(x,t) = ik exp(ikr iωt) r ( n f(n) ), (7) 1

exp(ikr iωt) ( ) E(x,t) = ikz 0 n (n f(n)), r (8) dp dω = Z 0k n f(n). (9) But the devil hides in details, or in our case in eq.eq. (6) for the direction-dependent amplitude f(n): For a short antenna of size λ, we may expand the amplitude f(n) into a series of electric and magnetic multipoles, but for longer antennas such expansion does not work. Instead, we need to evaluate the integral (6) in all its awful glory... Example: Center-Fed Linear Antenna As an example of a large antenna, consider a center-fed linear antenna of size L λ: generator I 0 e iωt I 0 e iωt L For simplicity, let s assume the antenna s diameter is negligibly thin compared to the antenna s length, so we may approximate the current in the antenna as J(x,y,z,t) = I(z)δ(x)δ(y)exp( iωt)n 3 (10) where n 3 is a unit vector along the z axis direction. For such a current, eq. (6) becomes f(n x ) = n 3 +L/ L/ dzi(z) exp ( ikn x z = ikzcosθ ). (11) Next consider the z dependence of the current I(z) in the antenna. To calculate this current from the first principles, we need equations and boundary conditions for the EM fields

immediately outside the antenna, which we should convert into an equation of the I(z), and then solve that equation. I shall address some of these issues at the end of these notes. But for the moment, let me simply assume that the current or rather the disturbances of the current propagates along the antenna at the speed of light. Consequently, the harmonically oscillating current forms a standing wave obeying ( d dz + k ) I(z) = 0, (1) except at z = 0 where the antenna has a small gap. Since the current must vanish at the two ends of the antenna, the standing wave has form I(z) = Î sin ( kl k z ) (13) where the amplitude Î is related to the feed current I 0 as Î = I 0 sin(kl/). (14) Here is the plot for L = 5 4 λ: z I(z) 3

For this current profile I(z), the integral (11) becomes f(n) = n 3 = n 3 +L/ L/ +L/ 0 dzîsin( 1 kl k z ) exp( ikcosθ z) dzîsin( 1 kl kz ) ( ) exp( ikcosθ z) + exp(+ikcosθ z) (15) where exp( ikcosθ z) + exp(+ikcosθ z) = cos(kcosθ z) (16) and sin ( 1 kl kz ) cos(kcosθ z) = sin ( 1 kl kz + kcosθ z ) + sin ( 1 kl kz kcosθ z ). (17) Plugging these formulae back into the integral (15), we obtain f(n) = În 3 where = În 3 +L/ 0 dz (sin ( 1 kl k(1 cosθ) z ) + sin ( 1 kl k(1+cosθ) z )) [ cos ( 1 kl k(1 cosθ) z ) k(1 cosθ) + cos( 1 kl k(1+cosθ) z ) k(1+cosθ) ] z=+l/ z=0 (18) atz = + 1L, cos( 1 kl k(1 cosθ) z ) = cos ( 1 kl 1 kl± 1kLcosθ) = cos( 1 klcosθ ) (19) while at z = 0, cos ( 1 kl k(1 cosθ) z ) = cos ( 1 kl). (0) 4

Consequently, on the last line of eq. (18), [ ] z=+l/ z=0 = cos ( 1 klcosθ ) ( ) 1 k(1 cosθ) + 1 k(1+cosθ) cos ( 1 kl ) ( ) 1 k(1 cosθ) + 1 k(1+cosθ) ( = cos ( 1 klcosθ ) cos ( 1 kl )) k(1 cos θ), (1) and therefore f(n) = În ( 3 πksin cos ( 1 θ klcosθ ) cos ( 1 kl )). () that Note that the direction of this vector amplitude f is n 3, the direction of the antenna, so nx f(n x ) = f(nx ) sin θ. (3) Consequently, plugging the amplitude () into eq. (9) for the EM power radiated in the direction n x = (θ,φ), we arrive at dp dω = Z 0 Î sin θ ( cos ( 1 kl cosθ ) cos ( 1 kl )), (4) or in terms of the feed current I 0 rather than the standing wave amplitude Î, dp dω = Z 0 I 0 sin ( 1 kl) ( ( cos 1 kl cosθ ) cos ( 1 kl ) ). (5) sinθ The second factor in this formula gives us the angular dependence of the power radiated by the linear antenna. Let s consider a few common examples in some detail. 5

A short dipole antenna, L λ. For such a short dipole cos ( 1 kl cosθ ) cos ( 1 kl ) (kl) 8 (1 cos θ) (6) and therefore dp dω ( 1 cos ) θ = sin θ. (7) sinθ Graphically, this direction-dependence of the radiated power is depicted on the radiation pattern diagram z axis (x,y) plane For the short dipole antenna, this radiation pattern diagram has a single main lobe centered at the horizontal plane, but this main lobe is rather thick, so a lot of radiation goes in directions well above or well belows the horizontal. A half-wavelength antenna, L = 1 λ, thus 1 kl = π/. This is a very common antenna length for the UHF radio waves or microwaves. For such an antenna and hence cos ( 1 kl cosθ ) cos ( 1 kl ) = cos ( π cosθ ) 0 (8) ( dp π dω cos cosθ ) sin. (9) θ The radiation pattern diagram for the half-wavelength antenna is the thick red line on For a general antenna, such a diagram should be three-dimensional, but thanks to the axial symmetry of the linear antenna, a D diagram like the above is sufficient. 6

this plot: z axis (x,y) plane The thin blue line here is for comparison purposes: it s the radiation pattern diagram for a short-dipole antenna. We see that both radiation patterns have a single main lobe centered in the horizontal plane, bot for the half-dipole antenna this main lobe is not quite as thick as for the short-dipole antenna. Full wavelength antenna, L = λ and hence 1 kl = π. This is also a common antenna length for the UHF radio waves and microwaves. For the full-wavelength antenna cos ( 1 kl cosθ ) cos ( 1 kl ) = cos(π cosθ) + 1 = cos ( π cosθ ), (30) so the direction-dependence of the fill-wavelength antenna s radiation is ( dp π dω cos4 cosθ ) sin. (31) θ 7

The radiation patter diagram for the full wavelength antenna is z axis (x,y) plane Again, we have a single main lobe centered in the horizontal plane, but this time it s quite thinner than for the short-dipole antenna. Double wavelength antenna, L = λ, thus 1 kl = π. For this antenna cos ( 1 kl cosθ ) cos ( 1 kl ) = cos(π cosθ) 1 = sin (π cosθ) (3) = 8sin ( π cosθ ) cos ( π cosθ ) and therefore ( dp π dω cos4 cosθ ) sin 4 ( π cosθ) sin. (33) θ The radiation pattern diagram for the double-wavelength antenna is 8

z axis (x,y) plane Surprisingly, there aremajor lobes, onecentered at θ = 58 andthe other at θ = 1 i.e., 3 above or below the horizontal, while no radiation is emitted in the horizontal or vertical direction. For the longer antennas, L = nλ for n = 3,4,6,10, let me skip the formulae and simply plot the radiation pattern diagrams. L = 3λ: z axis (x,y) plane 9

Note diagonal major lobes and a minor horizontal lobe. L = 4λ: z axis (x,y) plane This time, we have two major lobes at steep angles from the horizontal, plus minor lobes at shallower angles. L = 6λ: z axis (x,y) plane 10

L = 10λ: z axis (x,y) plane In general, the radiation pattern diagram for an antenna of length L = nλ for some integer n has n lobes centered at angles θ which obtain from the transcendental equation πntan ( n π cosθ) = cosθ sin θ πnctan ( n π cosθ) = + cosθ sin θ for even n, for odd n. (34) Among the n lobes, the two major lobes are much larger than the other N minor lobes. In terms of their central θ angles, the major lobes corresponds to the largest and the smallest solutions of eqs. (34). In other words, the two major lobes are closest to the vertical direction of the antenna itself. 11

Now let s turn our attention from the directions in which the EM power is radiated by a long linear antenna to the net power radiated by it, P net = d Ω dp dω = Z 0 I0 sin (kl/) d Ω ( ( cos 1 kl cosθ ) cos ( 1 kl ) ). (35) sinθ Formally, the integral here evaluates to P net = Z 0 I 0 sin (kl/) 4cos (kl/) Cin(kL) cos(kl) Cin(kL). (36) sin(kl) Si(kL) + sin(kl) Si(kL) where Si(x) def = x 0 sin(t) t dt and Cin(x) def = x 0 1 cos(t) t dt (37) are the sine and cosine integral function. You may look them up at this Wikipedia page, and if you are interested, read the references cited there. Alas, the class time is too short to spend it on studying these functions, so there is not much we can do with eq. (36) as it is. Alternatively, I could have calculated the integral (35) numerically, but I doubt a Fortran code would be any more meaningful to you than eq. (36). Instead, let me simply copy the plot from the Wikipedia page on dipole antennas: 1

(38) The black line on this plot shows the input resistance R in i.e., the real part of the input impedance Z in of a center-fed linear antenna as a function of L/λ. The net power radiated by the antenna is related to this resistance as P net = I 0 R in. (39) Note very large input resistance for L λ and L λ, or more generally whenever L/λ is close to an integer. This happens because for L = λ an integer, the standing current wave in the linear antenna has a node at the center, so for a center-fed antenna the input current I 0 becomes very small compared to the standing wave amplitude Î. Consequently, we get a finite radiated power for a very small input current, which translates into a very large input resistance I 0. The blue line on the plot (38) shows the input reactance X in. i.e., the imaginary part of the input impedance Z in. The reactance is unrelated to the radiated power, so it cannot be obtained from our analysis in these notes. Instead, one has to solve the boundary conditions 13

of the EM fields just outside the antenna, extract the voltage between the two halves of the antenna from the solution, and then relate it to the input current I 0 to get the complex input impedance Z in = V/I 0. The reactance Im(Z in happens to depend logarithmically on the antenna s diameter a, so even for a very small diameter there is a notable difference between antennas with a = 10 3 λ and a = 10 4 λ, etc., etc. Here is a plot of reactances as functions of L/λ (in the neighborhood of (L/λ) 0.5) for different antenna diameters: (40) Curiously, for antenna of length exactly 1 λ, the reactance seems to be diameter-independent Z in = (73 + j4.5) Ω. (41) 14

Antenna as a Boundary Problem Earlier in these notes I have assumed that the current in the antenna propagates at the speed oflight andthereforeformsa1dstanding wave. Nowlet s tryamoreaccurate analysis of the antenna current I(z). For simplicity, let s assume that (1) the antenna is made from a round wire of uniform diameter a much smaller than the antenna s length or the wavelength; () the antenna wire is a perfect conductor, σ ; (3) the antenna is surrounded by the vacuum. In light of the third assumption, everywhere outside the antenna E = ic ω ic ic ( B = A = ( A) A ). (4) ω ω Moreover, since the current in the antenna flows only in the ±z direction, the vector potential A(x) also points in the ±z direction, A(x) = A z (x)n 3, hence A = A z z (43) and therefore E z (x) = ic ω ( z ) A z (x). (44) Furthermore, the vector potential obeys the wave equation A z = k A z, so eq. (44) for the vertical electric field becomes E z (x) = ic ω ( z + k ) A z (x) = 0. (45) Eq. (45) should hold everywhere outside the antenna, and in particular right next to the antenna s surface. But for a perfectly conducting antenna, the electric field immediately outside its surface should be normal to the surface, thus in cylindrical coordinates (s, φ, z) Or rather, E z (s = a,z) = 0. (46) E z (s = a,z) = Vδ(z) (47) where V is the voltage amplitude between the two halves of the antenna. Comparing eqs. (47) 15

and (45), we obtain an equation for the vector potential immediately outside the antenna: at s = a and 1 L z +1 L, ( Solving this 1D differential equation gives us z + k ) A z (s,z) = iωv c δ(z). (48) A z (s = a,z) = αcos(kz) + βsin(kz) iv c sin(k z ) (49) for some constant coefficients α and β. In light of the upside-down symmetry of the centerfed antenna, the β coefficient should vanish, but the α coefficient remains unknown at this stage, thus for z 1 L, A z((s = a,z) = αcos(kz) iv c sin(k z ). (50) At the same time, everywhere outside the antenna and in particularly right outside the antenna s surface A(x) = µ 0 d 3 y eik x y J(y). (51) x y In a perfectly conducting antenna, the current flows only on the surface due to the skin effect, thus J(s,φ,z ) = I(z ) πa δ(s a)n 3. (5) Plugging this current into eq. (51), we find that the vector potential immediately outside the antenna s surface is A z (s = a,φ,z) = µ 0 8π +L/ L/ dz I(z ) π 0 dφ exp(ikr(z z,φ φ )) R(z z,φ φ ) (53) where R(z z,φ φ ) is the distance between the two points (s = a,φ,z) and (s = a,φ,z ) 16

on the antenna s surface, R(z z,φ φ ) = (z z ) + a (1 cos(φ φ )). (54) In the context of the integral (53), let s define the kernel K(x x ) = π 0 dφ π ( exp ik ) (z z ) +a (1 cosφ) (z z ) +a (1 cosφ) In terms this kernel, the vector potential immediately outside the antenna is. (55) A z (s = a,z) = µ 0 +L/ L/ dz I(z ) K(z z ). (56) At the same time, the same vector potential immediately outside the antenna is given by eq. (50), which gives us an integral equation for the current I(z): for z 1 L, µ 0 +L/ L/ dx I(z ) K(z z ) = αcos(kz) iv c sin(k z ) (57) for some constant α. Solving this integral equation is a major pain in the, so let s no go there in this class. Instead, let mesimply statethatinthelimit ofanextremely thin antenna, thesolution becomes the standing wave I(z) Îsin( 1 kl k z ). (58) However, the relative corrections to this formula scale like 1/ ln(ka) rather than any positive power of the radius a, so even for a 10 4 λ, the standing wave approximation could be off by 10 15%. 17